Question

A 0.86 kg rock is projected from the edge of the top of a building with an initial velocity of 8.65 m/s at an angle 46◦ above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 13.7 m from the base of the building. Building 13.7 m 46◦ 8.65 m/s h How tall is the building? Assume the ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s2 . Answer in units of m.

Answer 1

Answer:

Explanation:

Given

mass of rock=0.86 kg

initial velocity=8.65 m/s

launch angle$=46^{\circ}$

horizontal distance=13.7 m

let $u_x$ be the horizontal velocity given by $8.65\cos 46$

and $u_y$ be the vertical velocity $= 8.65\sin 46$

$R=u_x\times t$

$13.7=8.65\cos 46\times t$

t=2.279 s

and rock will start covering building height after the completion of its Projectile motion

time taken by stone for zero vertical displacement

$T=\frac{2u\sin \theta }{g}$

$T=\frac{2\times 8.65\times \sin 46}{9.8}=1.269 s$

thus time taken by rock to cover building height is 2.279-1.269=1.009 s

Let h be the height

$h=u\sin \theta t+\frac{gt^2}{2}$

h=6.27+4.98=11.25 m

Answer 2

Answer:

Height of the building = 11.4 m

Explanation:

As we know that the stone is projected at an angle 46 degree with speed 8.65 m/s

so the two components of the speed is given as

$v_x = 8.65 cos46$

$v_x = 6 m/s$

vertical component of the speed is given as

$v_y = 8.65 sin46$

$v_y = 6.22 m/s$

now we know that the ball strike at horizontal distance of 13.7 m

so we will have

$x = v_x t$

$13.7 = 6 t$

$t = 2.28 s$

now we know that in vertical direction ball will move under uniform gravity so we can use kinematics

$y = v_y t + \frac{1}{2}at^2$

$y = 6.22(2.28) - \frac{1}{2}(9.81)(2.28^2)$

$y = -11.4 m$

Height of the building = 11.4 m